derivative f(t)=(2t+1)/(t+3)

Lösung

ableitung von

f (t) = \frac{2 t + 1}{t + 3}

\frac{5}{( t + 3 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (\frac{2 t + 1}{t + 3})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( 2 t + 1 ) ( t + 3 ) - \frac{d}{d t} ( t + 3 ) ( 2 t + 1 )}{( t + 3 ) ^{2}}

\frac{d}{d t} (2 t + 1) = 2

\frac{d}{d t} (t + 3) = 1

= \frac{2 ( t + 3 ) - 1 \cdot ( 2 t + 1 )}{( t + 3 ) ^{2}}

Vereinfache

\frac{2 ( t + 3 ) - 1 \cdot ( 2 t + 1 )}{( t + 3 ) ^{2}} : \frac{5}{( t + 3 ) ^{2}}

= \frac{5}{( t + 3 ) ^{2}}

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